Show that the reaction : CO(g) + 1/2O_(2)(g) to CO_(2)(g) at 300 K is spontaneous and exothermic, when standard entropy change is -0.094 kJ"mol"^(-1).The standard Gibbs energies of formation of CO_(2) and CO are -394.4 and -137.2 kJ"mol"^(-1) respectively.

Show that the reaction : CO(g) + 1/2O_(2)(g) to CO_(2)(g) at 300 K is

1.	Show that the reaction : CO(g) + 1/2O_(2)(g) to CO_(2)(g) at 300 K is spontaneous and exothermic, when standard entropy change is -0.094 kJ"mol"^(-1).The standard Gibbs energies of formation of CO_(2) and CO are -394.4 and -137.2 kJ"mol"^(-1) respectively.
Answer» Solution :`Delta_(R)G^(@) = Delta_(f)G^(@)("products") - Delta_(f)G^(@)("REACTANTS")` ` = [Delta_(f)G^(@(CO_(2))] - [Delta_(f)G^(@)(CO) + 1/2Delta_(f)G^(@)(O_(2))]` `= -394.4 - [-137.2 + 0]` ` = -257.2 kJ` Since `Delta_(r)G^(@)` is negative, the REACTION is spontaneous. Now `DELTAG^(@) = DeltaH^(@) -TDeltaS` ` -257.2 = DeltaH^(@) -[300 xx(-0.094)]` `therefore DeltaH^(@) = -257.2 -28.2 = -285.4 kJ` Since `DeltaH^(@)` is negative, the reaction is exothermic.